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I have a question that gives two concentrations and asks for the mass of $\ce{HCl}$ formed by the reaction.

$$\ce{H2SO4 + NaCl ->Na2SO4 + HCl}$$

I have two concentrations:

$\pu{250 mL}$ of $\pu{4.00 M}$ $\ce{H2SO4}$, and $\pu{250 mL}$ of $\pu{1.00 M}$ $\ce{NaCl}$.

Here is the balanced reaction equation:

$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.

However, how do I add those two concentrations? I assume the mixture will be at least $\pu{500 mL}$, but how do I add the molarity?

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Your working equation is correct. $$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

  • Find the amount of substance of protons of each solution via $n=c\cdot V$.

$n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

  • What is the limiting agent?

Chlorine

  • How many moles of hydrogen chloride can only be formed?

$n(\ce{HCl})=0.25~\mathrm{mol}$

  • Calculate the mass of hydrogen chloride via $m = n\cdot M$

$M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$

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  • $\begingroup$ Martin, you are a formatting genius. $\endgroup$
    – Dissenter
    Commented Jun 26, 2014 at 16:04
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    $\begingroup$ @Dissenter As a child I loved playing in a sandbox... so I enjoy that now and again too: Check this out! $\endgroup$ Commented Jun 27, 2014 at 2:09
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You're right; you can't directly add the two concentrations since these concentrations are specific to a certain volume of solvent.

If you want to recalculate concentrations of each system component in a combined solution, you need to find the moles of each system component in this combined solution, and then divide by the total solution volume.

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  • $\begingroup$ So does that become 500mL of 2.500 M solution? $\endgroup$
    – Mark
    Commented Jun 26, 2014 at 3:50
  • $\begingroup$ That is not quite correct, as you have a limiting agent present. $\endgroup$ Commented Jun 26, 2014 at 6:16

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